Question

Find log(a³-3a²b+3ab²-b²)
[ base ( a²-2ab+b²) ].

Answer 1

As we know that :

$( {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} ) = {(a - b)}^{3} \\ \\ an d \: \: ( {a}^{2} - 2ab + {b}^{2} ) = {(a + b)}^{2}$

Also, as we know that :

$log_{n}(m) = \frac{ log(m) }{ log(n)} \\ \\ here \: m = ( {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} ) = {(a - b)}^{3} \\ \\ and \: \: n = ( {a}^{2} - 2ab + {b}^{2} ) = {(a - b)}^{2}$

Now, according to the question :

$\frac{ log {(a - b)}^{3} }{ log {(a - b)}^{2} } \\ \\ = > \frac{3 log(a - b) }{2 log(a - b) } \\ \\ As \: we \: know \: that :\: log {m}^{n} = nlogm \\ \\ = > \frac{3}{2}$

Answer 2

(a + b)^3 can be written as,

(a + b) * (a + b) * (a + b)

Let's multiply the first two terms first.

[ (a +b) * (a + b) ] * (a + b)

[ a*a + a*b + b*a + b*b] * (a + b)

(a^2 + ab + ab + b^2) * (a + b)

(a^2 + 2ab + b^2) * (a + b)

Note :The term in the first parenthesis is the formula of (a + b) ^2 = (a + b) * (a + b)

Let's proceed further!

Let's multiply the the terms in two parentheses.

=> ( a^2 * a + a^2 * b +

2ab * a + 2ab * b +

b^2 * a + b^2 * b )

=> (a^3 + ba^2 + 2ba^2 + 2ab^2 + ab^2 + b^3)

=> (a^3 + 3ba^2 + 3ab^2 + b^3)

We can rewrite this formula much further. If you see, '3ab’ is common in 2nd and 3rd terms. So, let's take it out as common.

=>[a^3 + 3ab (a + b) + b^3]

Therefore, we can say,

(a + b)^3 = a^3 + 3ab (a + b) + b^3

Hope this helps!